Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $q = \dfrac{-2t - 18}{t^2 - 81} \div \dfrac{t - 4}{2t^2 - 14t + 24} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{-2t - 18}{t^2 - 81} \times \dfrac{2t^2 - 14t + 24}{t - 4} $ First factor out any common factors. $q = \dfrac{-2(t + 9)}{t^2 - 81} \times \dfrac{2(t^2 - 7t + 12)}{t - 4} $ Then factor the quadratic expressions. $q = \dfrac {-2(t + 9)} {(t + 9)(t - 9)} \times \dfrac {2(t - 4)(t - 3)} {t - 4} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac {-2(t + 9) \times 2(t - 4)(t - 3) } { (t + 9)(t - 9) \times (t - 4)} $ $q = \dfrac {-4(t - 4)(t - 3)(t + 9)} {(t + 9)(t - 9)(t - 4)} $ Notice that $(t + 9)$ and $(t - 4)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {-4(t - 4)(t - 3)\cancel{(t + 9)}} {\cancel{(t + 9)}(t - 9)(t - 4)} $ We are dividing by $t + 9$ , so $t + 9 \neq 0$ Therefore, $t \neq -9$ $q = \dfrac {-4\cancel{(t - 4)}(t - 3)\cancel{(t + 9)}} {\cancel{(t + 9)}(t - 9)\cancel{(t - 4)}} $ We are dividing by $t - 4$ , so $t - 4 \neq 0$ Therefore, $t \neq 4$ $q = \dfrac {-4(t - 3)} {t - 9} $ $ q = \dfrac{-4(t - 3)}{t - 9}; t \neq -9; t \neq 4 $